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Sidana [21]
2 years ago
7

In an electroplating apparatus, where does the reduction reaction occur?

Chemistry
1 answer:
Ludmilka [50]2 years ago
6 0

Answer:

on the surface of the cathode

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Which of these is something whose construction MOST LIKELY would be determined by the government? A) An airport B) A golf club C
Amiraneli [1.4K]

Answer:

I would say an airport

Explanation:

why because it is ran by the government

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Write the structural formula for a primary, a secondary, and a tertiary constitutionally isomeric alcohol of the molecular formu
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The structures of all isomeric alcohols of molecular formula C5H12O along with their IUPAC names are as shown.

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2-pentanol -

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3 years ago
1. Interpret the following equation using moles, molecules, and volumes (assume STP). Compare the mass of the reactants to the m
blsea [12.9K]
1. In this reaction, 2 moles of nitrogen gas reacts with 3 moles of oxygen gas to give 2 moles of N2O3 gas. 2 nitrogen molecules react with 3 oxygen molecules to give 2 N2O3 molecules.  Under STP, one mole of an ideal gas occupies a volume of 22.4 liters. So in this reaction, 44.8 liters of nitrogen gas reacts with 67.2 liters of oxygen gas to give 44.8 liters of N2O3 gas.  The total mass of the reactants (N2 and O2) is the same as the total mass of the product (N2O3). This is called mass balance of a chemical reaction.


2. According to the chemical reaction, 3 moles of chlorine gas produces 2 moles of iron(III) chloride.  So, to produce 1 moles of iron(III) chloride, 3/2 (1.5) moles of chlorine gas is required.  Therefore, to produce 14 moles of iron(III) chloride, 14 x 1.5 = 21 moles of chlorine is needed.
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4 years ago
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For the readion 2Na + Ch> 2NaCl, how many grams of Ch are required to read completely with 450 g Na
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I think it might be 3.54g
3 0
3 years ago
Enough of a monoprotic acid is dissolved in water to produce a 0.0165 m solution. the ph of the resulting solution is 2.68. calc
Dmitrij [34]
Answer is: Ka for the monoprotic acid is 3.03·10⁻⁴.
<span>
Chemical reaction: HA(aq) </span>⇄ A⁻(aq) + H⁺<span>(aq).
c(monoprotic acid) = 0.0165 M.
pH = 2.68.
[A</span>⁻] = [H⁺] = 10∧(-2.68).<span>
[A</span>⁻] = [H⁺] = 0.00209 M; equilibrium concentration.<span>
[HA] = 0.0165 M - 0.00209 M.
[HA] = 0.0144 M.
Ka = [A</span>⁻]·[H⁺] / [HA].<span> 
Ka = (0.00209 M)² / 0.0144 M.
Ka = 0.000303 M = 3.03·10</span>⁻⁴<span> M.</span>
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